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I studied first four chapters of Serre's book on linear representations of finite groups, which talks about irreducible representations and their characters, canonical decomposition of a representation, induced representations and their characters (along with obvious big theorems: Schur's lemma and Maschke's theorem).

Then I took the group $SL(2,3)$, a non-abelian group of order $24$, and moved for constructing its representations/characters. But, I could not complete the character table: I found irreducible characters of degree $1,1,1,3$ simply by pulling them via quotient modulo center $\pm I$.

I thought to get remaining by induction; but then after some hitting, I reached to conclusion that it has no subgroups of index $2$.

Then, how should I proceed to construct at least one or all the three irreducibles of degree $2$? (Should I get familiar with some further topics of Serre for this? )

(The existence of three irreducibles of degree $2$ can be guaranted from the rule that $24=1^2+1^2+1^2+3^2+ \sum_i n_i^2$ with $n_i>1$.)

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    $\begingroup$ Surely the character table can be completed using the orthogonality relations? $\endgroup$ Commented Oct 13, 2022 at 10:10
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    $\begingroup$ Inducing a degree $2$ charatcer of $Q_8$ up to $G={\rm SL}(2,3)$ gives the sum of the three degree $2$ characters. Inducting a degree $1$-character of $C_6$ with kernel of order $3$ to $G$ gives the sum of two algebraically conjugate degree $2$ characters. That should be enough information to deduce them together with the othogonality relations. $\endgroup$
    – Derek Holt
    Commented Oct 13, 2022 at 11:26

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I understand the 2D complex irrep best with quaternions.

Every quaternion looks like $a+b\mathbf{i}+c\mathbf{j}+d\mathbf{k}$, in other words the formal sum of a scalar and a 3D vector. The product of any two of $\mathbf{i},\mathbf{j},\mathbf{k}$ in cyclic order is the third, and in reverse order is the opposite. If you check the multiplication table, then, you find if $\mathbf{u},\mathbf{v}$ are 3D vectors then the scalar and vector components (aka real and imaginary parts) of the quaternion product $\mathbf{uv}$ are $-\mathbf{u}\cdot\mathbf{v}+\mathbf{u}\times\mathbf{v}$ (since it is true for basis vectors). This implies the square roots of $-1$ are the unit vectors, and Euler's formula $\exp(\theta\mathbf{v})=\cos(\theta)+\sin(\theta)\mathbf{v}$ for unit vectors $\mathbf{v}$ follows. Furthermore, every quaternion has a polar form $r\exp(\theta\mathbf{v})$ with $r\ge0$ its magnitude, $\mathbf{v}$ a unit vectors, and $0\le\theta\le\pi$ convex (which is unique for nonreal quaternions, and $\theta=0$ or $\pi$ with $\mathbf{v}$ arbitrary for nonzero reals). If $\mathbf{u}$ is a unit vector, we can extend to an ordered orthonormal basis $\{1,\mathbf{u},\mathbf{v},\mathbf{w}\}$ of the quaternion algebra. If we write $L_q(x)=qx$ and $R_q(x)=xq$ with respect to this basis, where $q=\exp(\theta\mathbf{u})$, we see that the conjugation $C_q=L_q\circ R_q^{-1}$ given by $C_q(x)=qxq^{-1}$ restricts to a rotation of 3D vectors around the oriented axis by an angle of $2\theta$. In this way, $q\mapsto C_q$ is a double covering (i.e. $2$-to-$1$ group homomorphsim) $S^3\to\mathrm{SO}(3)$, where $S^3$ is the group of unit quaternions.

If $T\cong A_4$ is the tetrahedral group (of order $12$), its preimage under $S^3\to\mathrm{SO}(3)$ is the binary tetrahedral group $2T$ of order $24$. It forms the vertices of a $24$-cell, namely $\{\pm1,\pm\mathbf{i},\pm\mathbf{j},\pm\mathbf{k},\frac{1}{2}(\pm1\pm\mathbf{i}\pm\mathbf{j}\pm\mathbf{k})\}$. The group is a semidirect product $2T=Q_8\rtimes C_3$ with $C_3$ generated by any $3$rd root of unity $\frac{1}{2}(-1\pm\mathbf{i}\pm\mathbf{j}\pm\mathbf{k})$, with polar form $\omega=\exp(\frac{2\pi}{3}\frac{\pm\mathbf{i}\pm\mathbf{j}\pm\mathbf{k}}{\sqrt{3}})$. Geometrically, we can place unit diameter $3$-spheres at all integer coordinates in 4D, which leaves just enough space to place unit diameter spheres centered at the hypercube centers to create an optimal sphere packing (the Hurwitz quaternions); the vertices of the spheres surrounding the central sphere forms the $24$-cell.

We can have $2T\curvearrowright\mathbb{H}$ act from the left and complex scalars from $\mathbb{C}=\{a+b\mathbf{i}\}$ act from the right (so it is a "right" vector space). Picking a basis $\{1,\mathbf{j}\}$ for $\mathbb{H}$ as a complex vector space allows us to write the right $\mathbb{C}$-linear transformations from $2T$ as $2\times2$ complex matrices. This is a 2D complex irrep of $2T$. We can check the character table to see tensoring this 2D rep with the nontrivial 1D reps gives the two other 2D irreps.

If we mod these $2\times2$ matrices by $3$ (the smallest modulus where $\pm1$ are distinct), we get $\mathrm{SU}_{1,1}(\mathbb{F}_{3^2})$, where $\mathbb{F}_{3^2}=\mathbb{F}_3(i)$, which is conjugate via the Cayley transform $[\begin{smallmatrix}1&-i\\1&i\end{smallmatrix}]$ to $\mathrm{SL}_2(\mathbb{F}_3)$. I elaborate a little more in my answer here (which shows the binary octahedral group $2O$ is a twisted version of $\mathrm{GL}_2\mathbb{F}_3$).

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  • $\begingroup$ There are three $2$-dimensional complex irreducible representations of ${\rm SL}(2,3)$, and it might be helpful to indicate which one of the three you are talking about here. Two of them are algebraically conjugate with Frobenius-Schur indicator $0$. The third is quaternionic with indicator $-1$, so I guess that is the one you are constructing. $\endgroup$
    – Derek Holt
    Commented Oct 14, 2022 at 8:05

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